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A metallic laboratory spring is typically 5.00 cmcm long and 0.150 cmcm in diameter and has 50 coils. Part A If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid? Express your answer to three significant figures and include the appropriate units.

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boffeemadrid

Answer:

[tex]1.11\times 10^{-7}\ H[/tex]

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]

l = Length of spring = 5 cm

d = Diameter = 0.15 cm

A = Area = [tex]\dfrac{\pi}{4}d^2[/tex]

N = Number of turns = 50

Inductance is given by

[tex]L=\dfrac{\mu_0L^2A}{l}\\\Rightarrow L=\dfrac{4\pi \times 10^{-7}\times 50^2\dfrac{\pi}{4}(0.15\times 10^{-2})^2}{0.05}\\\Rightarrow L=1.1103304951\times 10^{-7}\ H[/tex]

The self inductance is [tex]1.11\times 10^{-7}\ H[/tex]

onyebuchinnaji

The inductance of a solenoid that must be included in the model is 1.11 x 10⁻⁷ H.

The given parameters;

  • length of the spring, L = 5 cm = 0.05 m
  • diameter of the spring, d = 0.15 cm
  • radius of the spring, r = 0.075 cm = 0.00075 m
  • number of coils, N = 50

The area of the spring is calculated as follows;

[tex]A = \pi r^2\\\\A = \pi \times (0.00075)^2\\\\A = 1.77 \times 10^{-6} \ m^2[/tex]

The inductance of a solenoid is calculated as follows;

[tex]L = \frac{N^2 \times \mu_0 \times A}{l} \\\\L = \frac{(50)^2 \times (4\pi \times 10^{-7} )\times (1.77 \times 10^{-6})} {0.05} \\\\L = 1.11 \times 10^{-7} \ H\\\\[/tex]

Thus, the inductance of a solenoid that must be included in the model is 1.11 x 10⁻⁷ H.

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